Question

A manufacturing company makes two types of teaching aids A and B of Mathematics for Class X. Each type of A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30 respectively. The company makes a profit of Rs. 80 on each piece of type A and Rs. 120 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Formulate this as Linear Programming Problem and solve it. Identify the feasible region from the rough sketch.

Answer 1

	A(x)	B(y)	Total
Fabricating	9	12	180
Finishing	1	3	30

Let no. of pieces of type A and type B manufactured be x and y respectively.

Subject to the constraints :

9x + 12y ≤ 180

x + 3y ≤ 30

x ≥ 0

y ≥ 0

Maximise Z = 80 x + 120 y

9x + 12y = 180

x	0	20	12
y	15	0	6

x + 3y = 30

x	0	30	12
y	10	0	6

9x + 12y ≤ 180

The feasible region of the LPP is shaded in the fig, the corner points of the feasible region OAPR are (0, 0), (20, 0), (12, 6) and (0, 10)

These points have been obtained by solving the cor responding intersecting lines simultaneously The value of the objective function Z at corner points of the feasible region are given is the following table :

Point (x, y)	Z = 80x + 120y
0(0,0) A (20,0) P (12,6) B (0,10)	Z = 0 Z = 1600 Z = 1680 Z = 120

Clearly, Z is maximum at x = 12 and y = 6 The Maximum value of Z is Rs.1680.

Hence the manufacturer should manufactured 12 aids of A and 6 aids of B to obtain maximum profit under the given conditions.